问题
I need an algorithm (preferable in a Pascal-like language, but it the end doesn't really matter) that will make the "signal" (actually a series of data points) in left look like the one in right.
Signal origin:
The signal is generated by a machine. Oversimplifying the explanation, the machine is measuring the density of a liquid flowing through a transparent tube. So, the signal is nothing similar to an electrical signal (audio/radio frequency).
The data points could look like this: [1, 2, 1, 3, 4, 5, 4, 3, 2, 1, 13, 14, 15, 18, 23, 19, 17, 15, 15, 15, 14, 11, 9, 4, 1, 1, 2, 2, 1, 2]
What I need:
I need to accurately detect the 'peaks'. For this, I already have a piece of code but it is not working on poor signals as the one shown in the image below.
I think we can see this, as a signal that was accidentally passed through a low-pass filter, and now I want to restore it.
Notes:
There are 4 signals, but they are separated so they can be analyzed individually. So, it is enough to think about how to process just one of them.
After a peak, if the signal is not coming down fast enough we can consider that there are multiple peaks (you can best see that in the 'red' signal at the end of the series).
The advantage is that the whole series is available (so the signal is not in real time, it is already stored on file)!
[Edit by Spektre] I extracted Red sample points from the image
float f0[]={ 73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,72,71,69,68,66,64,62,58,54,49,41,33,25,17,13,15,21,30,39,47,54,59,62,64,66,67,68,69,70,71,71,72,72,72,71,71,70,69,68,67,66,65,63,62,60,56,51,45,37,29,22,18,18,22,28,33,35,36,35,32,26,20,15,12,15,20,26,31,35,37,36,34,30,25,22,22,27,33,41,48,55,60,63,66,67,68,69,70,71,72,72,73,73,73,73,73,73,72,71,70,69,67,65,63,60,55,49,40,30,21,13, 7,10,17,27,36,45,52,56,59,60,61,62,62,62,62,61,61,59,57,53,47,40,32,24,18,15,18,23,28,32,33,31,28,23,16,10, 6, 8,13,20,27,31,32,31,28,22,15,10, 6,10,16,23,30,34,36,36,34,29,24,20,19,24,30,37,44,51,56,59,61,62,63,64,64,64,65,64,64,62,60,57,53,48,43,38,36,39,43,49,54,59,63,66,68,69,70,71,72,72,73,73,73,73,73,73,73,73,73,73,73,73,73,73 };
float f1[]={ 55,58,60,62,64,66,67,68,68,69,69,70,71,72,72,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,73,72,72,72,72,72,72,72,72,72,72,72,72,72,72,72,73,73,73,72,72,72,72,71,71,71,71,71,71,70,70,69,68,67,66,64,63,61,60,59,57,55,52,49,46,43,40,37,35,34,33,32,32,33,34,36,37,39,41,43,45,47,50,52,55,57,59,60,61,61,62,62,62,62,61,61,60,58,57,55,53,51,49,48,46,44,42,40,38,35,32,30,29,28,27,27,26,26,26,25,25,24,23,23,23,24,24,25,25,26,26,26,27,28,29,31,33,35,38,40,41,43,44,46,48,50,53,55,57,59,60,61,62,63,64,64,65,65,64,63,61,59,57,54,52,50,47,45,42,39,37,34,32,31,30,30,30,31,32,34,36,37,39,40,41,42,43,44,44,44,44,43,42,41,40,38,36,34,32,30,28,26,25,24,23,22,21,20,18,17,17,17,17,18,18,18,18,18,18,18,18,18,18,19,19,19,19,20,20,21,23,24,25,26,26,26,27,28,29,31,34,36,37,38,40,41,43,45,47,48,49,50,51,51,51,50,49,49,48,48,47,47,47,47,47,47,48,60 };
const int n=sizeof(f0)/sizeof(f0[0]);
All the values need to be transformed:
f0[i] = 73.0-f0[i];
f1[i] = 73.0-f1[i];
To offset back from image... f0 is the original Red signal and f1 is the distorted Yellow one.
This is the closest I get to with first order FIR filter:
The upper half is plot of used FIR filter weights (editable by mouse so the FIR is hand drawed for fast weights find...). Bellow are the signal plots:
- Red original (Ideal) signal
f0 - Dark green measured signal
f1 Light Green FIR filtered Ideal signal
f2The FIR filter is just convolution where zero offset element is the last and here the weight values:
float fir[35] = { 0.0007506932, 0.0007506932, 0.0007506932, 0.0007506932, 0.0007506932, 0.0007506932, 0.0007506932, 0.0007506932, 0.0007506932, 0.003784107, 0.007575874, 0.01060929, 0.01591776, 0.02198459, 0.03032647, 0.04170178, 0.05686884, 0.06445237, 0.06900249, 0.07203591, 0.07203591, 0.0705192, 0.06900249, 0.06672744, 0.06217732, 0.05611049, 0.04928531, 0.04170178, 0.03335989, 0.02653471, 0.02046788, 0.01515941, 0.009850933, 0.005300813, 0.0007506932 };
So either there is also some higher degree of FIR or the weights need to be tweaked a bit more. Anyway this should be enough for the deconvolution and or fitting ... btw the FIR filter is done as follows:
const int fir_n=35; // size (exposure time) [samples]
const int fir_x=fir_n-1; // zero offset element [samples]
int x,y,i,j,ii;
float a,f2[n];
for (i=0;i<n;i++)
for (f2[i]=0.0,ii=i-fir_x,j=0;j<fir_n;j++,ii++)
if ((ii>=0)&&(ii<n)) f2[i]+=fir[j]*f0[ii];
回答1:
In my opinion, you should start with a pretty simple system identification and consecutive signal reconstruction. Also, I would recommend to implement your algorithm first in a mathematical prototyping tool like Matlab (commerical license) or Octave (free → https://www.gnu.org/software/octave/download.html). These tools provide an ease of signal processing no programming language like Pascal or Java could ever offer, no matter what library you use. After you successfully designed your algorithm with Matlab or Octave, then think about how to implement it with Pascal.
Lets assume the behaviour of the tube can be characterized by a linear time-invariant system (e.g. a linear lowpass filter). This is by no means guaranteed but a worthwhile approach (at least until it fails:) ). Following the same approach for non-linear and/or time-variant systems becomes pretty involved and you will need professional help to do this I would imagine.
If I understood your description correctly, you have access to both the input and output signals of the tube. If I am wrong and you don’t know the input signal, you might be able to apply some calibration signal first, whose characteristics you know, and record the output signal. Knowing input and output signals is a prerequisite for the following approach. Without both signals you can not approximate the impulse response h of the tube. After calculating an approximation of h, we can design an inverse filer called ge and eventually reconstruct the input from the output signal.
Here is the signal flow of a input signal x[n] passing your tube described by h producing the output signal y[n]. Taking y[n] and applying an inverse filtering operation described by ge we obtain xr[n]
x[n] →| h | → y[n] → | ge | → xr[n]
Take an input vector x of length N and a corresponding vector of y of the same length. Now you express the output y as a convolution of an input convolution matrix X (see the code below for its implementation) with the unknown impulse response of your system, i.e.
y = X * h
with the vector and matrix sizes y = N x 1, X = N x N and h = N x 1 You can calculate a least-squares approximation of the impulse response he, by calculating
he = inv(X'*X)*X' * y
where X' describes the transpose and inv() the matrix inverse of X. he represents a column vector of the identified impulse response of your tube which we obtained via a 1-D deconvolution. You can check how well the identification worked by calculating the output of your estimated system,
ye = X * he
and by comparing ye and y. Now, we try to reconstruct x from y and he. The reconstructed input vector xr is calculated by
xr = Ge * y
where Ge = inv(He) and He is the N x N convolution matrix of he. Here is some Octave code. Copy both functions in their own dedicated file (reconstruct.m and getConvolutionMatrix.m) and type "reconstruct.m" into the Octave command line to examine the outputs of the example. Please note the sample code works only for vector of odd length (N is odd). Play around with the size N of your vectors. This might help the approximation accuracy.
function [Ge] = reconstruct ()
x = [1 2 3]'; # input signal
# h = [1 3 2]'; # unknown impulse response
y = [5 11 13]'; # output signal
y = y + 0.001*randn(length(y),1) # add noise to output signal
Xm = getConvolutionMatrix(x)
Xps = inv(Xm'*Xm)*Xm';
he = Xps * y
He = getConvolutionMatrix(he);
Ge = inv(He);
# reconstructed impulse signal
xr = Ge*y
endfunction
function [mH] = getConvolutionMatrix(h)
h = h(:)';
hlen = length(h);
Nc = (hlen-1)/2;
mH= zeros(hlen, hlen);
hp = [zeros(1,Nc) h zeros(1,Nc)];
for c=1:hlen
for r=1:hlen
mH(r,c) = hp(r+hlen-c);
end
end
endfunction
来源：https://stackoverflow.com/questions/38330250/signal-enhancing-algorithm